If $I_n=\int x^n \cdot e^{c x} d x$ for $n \geq 1, \quad$ then $c \cdot I_n+n \cdot I_{n-1}$ is equal to
If $I_n=\int x^n \cdot e^{c x} d x$ for $n \geq 1, \quad$ then $c \cdot I_n+n \cdot I_{n-1}$ is equal to
- $x^n e^{c x}$
- $x^n$
- $e^{c x}$
- $x^n+e^{c x}$
Solution
Given that,
$
\begin{aligned}
& I_n=\int x^n \cdot e^{c x} d x \\
& I_n=\frac{e^{c x}}{c} \cdot x^n-\int \frac{e^{c x}}{c} \cdot n x^{n-1} d x \\
& \Rightarrow \quad I_n=\frac{e^{c x} \cdot x^n}{c}-\frac{n}{c} I_{n-1} \\
&
\end{aligned}
$
$\Rightarrow \quad c I_n+n I_{n-1}=e^{c x} \cdot x^n$
Asked in: AP EAMCET 2008
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